š Step 1: Problem Identification
What type of problem is this?
š¢ Math/Number Theory
š³ Tree Traversal
šÆ Two Pointers/Sliding Window
š Dynamic Programming
Attempts: 0
š” Hint: Think about maintaining a "window" of characters. When you encounter a duplicate, you need to shrink the window from one side. This technique is commonly used for substring problems.
ā
Explanation: This is a classic sliding window problem. We need to find the longest substring without repeating characters, which is perfectly suited for the two-pointers/sliding window technique where we expand and contract a window based on conditions.
š§© Step 2: Logical Intuition
Arrange the pseudocode blocks in the correct order:
Available Blocks:
Return maxLength
Initialize left = 0, maxLength = 0
For each right pointer from 0 to n-1
While character at right exists in current window
Move left pointer forward
Update maxLength if current window is larger
Solution Order:
Attempts: 0
š” Hint: Start with initialization, then think about expanding the window with the right pointer. When you find a duplicate, you need to shrink the window by moving the left pointer. Don't forget to track the maximum length!
ā
Explanation: The correct order implements a sliding window: initialize variables, expand window with right pointer, contract when duplicates found by moving left pointer, track maximum length, and return result.
ā” Step 3: Algorithm Selection
Which approach is most suitable for this problem?
Brute Force - Check all substrings
Sliding Window with HashSet
Binary Search
Divide and Conquer
Attempts: 0
š” Hint: We need to efficiently track which characters we've seen and be able to quickly check for duplicates. What data structure is perfect for O(1) lookups?
ā
Explanation: Sliding Window with HashSet is optimal for this problem. It maintains a window of unique characters and efficiently expands/contracts it, achieving O(n) time complexity compared to O(nĀ²) brute force.
ā±ļø Step 4: Complexity Analysis
function lengthOfLongestSubstring(s) {
let maxLen = 0;
for (let i = 0; i < s.length; i++) {
let seen = new Set();
for (let j = i; j < s.length; j++) {
if (seen.has(s[j])) break;
seen.add(s[j]);
maxLen = Math.max(maxLen, j - i + 1);
}
}
return maxLen;
}
What's the time complexity of this code?
O(n) - Linear
O(nĀ²) - Quadratic
O(n log n) - Linearithmic
O(nĀ³) - Cubic
Attempts: 0
š” Hint: Count the nested loops! The outer loop runs n times, and for each iteration, the inner loop can run up to n times in the worst case.
ā
Explanation: This brute force approach has O(nĀ²) complexity because of the nested loops. For each starting position (outer loop), we check all possible endings (inner loop). The optimal sliding window solution would be O(n).
šÆ Step 5: Output Prediction
Given the input string "pwwkew", what will be the output?
4 (substring "pwke")
3 (substring "wke")
2 (substring "pw")
6 (entire string)
Attempts: 0
š” Hint: Remember that a substring must be contiguous! Trace through "pwwkew" and find the longest contiguous part without any repeating characters.
ā
Explanation: For "pwwkew", the longest substring without repeating characters is "wke" with length 3. Note that "pwke" is a subsequence, not a substring (must be contiguous).
š Step 6: Edge Case Detection
function lengthOfLongestSubstring(s) {
let left = 0, maxLen = 0;
let charMap = new Map();
for (let right = 0; right < s.length; right++) {
if (charMap.has(s[right])) {
left = charMap.get(s[right]) + 1;
}
charMap.set(s[right], right);
maxLen = Math.max(maxLen, right - left + 1);
}
return maxLen;
}
This code has a bug! Enter a test case that will make it fail:
Attempts: 0
š” Hint: Think about what happens when the left pointer would move backward. Try a string where you see a character again, but it appeared before the current left position.
ā
Explanation: Try "abba" or "tmmzuxt". The bug occurs when the left pointer moves backward. For "abba": when we see the second 'a', left jumps to position 1, but should stay at 2.
š Step 7: Code Debugging
Click on the line that contains the bug:
function lengthOfLongestSubstring(s) {
let left = 0, maxLen = 0;
let charMap = new Map();
for (let right = 0; right < s.length; right++) {
if (charMap.has(s[right])) {
left = charMap.get(s[right]) + 1;
}
charMap.set(s[right], right);
maxLen = Math.max(maxLen, right - left + 1);
}
return maxLen;
}
Attempts: 0
š” Hint: Look for the line where the left pointer is updated. The bug is that the left pointer can move backward, which breaks the sliding window property.
ā
Explanation: The bug is in line "left = charMap.get(s[right]) + 1;" because it can move the left pointer backward. The fix is to use Math.max(left, charMap.get(s[right]) + 1).
āļø Step 8: Code Completion
Fill in the blanks to fix the bug:
function lengthOfLongestSubstring(s) {
let left = 0, maxLen = 0;
let charMap = new Map();
for (let right = 0; right < s.length; right++) {
if (charMap.has(s[right])) {
left = (charMap.get(s[right]) + 1, );
}
charMap.set(s[right], right);
maxLen = Math.max(maxLen, right - left + 1);
}
return maxLen;
}
Attempts: 0
š” Hint: You need to ensure the left pointer never moves backward. What function can help you choose the maximum between two values?
ā
Explanation: The correct code is "left = Math.max(left, charMap.get(s[right]) + 1);" - this ensures the left pointer never moves backward, which maintains the sliding window property.